Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool vis[100010];
vector<int> cn[100010];
vector<int> D;
vector<vector<int> > T;
vector<vector<vector<int> > > BM;
vector<vector<int> > c;
void dfs(int x) {
  vis[x] = true;
  for (int i = 0; i < cn[x].size(); i++) {
    int v = cn[x][i];
    if (vis[v]) continue;
    D[v] = D[x] + 1;
    T[v][0] = x;
    BM[v][0] = c[x];
    dfs(v);
  }
}
void gm(vector<int>& a, const vector<int>& b) {
  int mo = min(10, int(a.size() + b.size()));
  vector<int> c = a;
  int pb = 0;
  int pc = 0;
  a.clear();
  for (int i = 0; i < mo; i++) {
    if (pc == c.size() && pb == b.size()) break;
    if (pb == b.size() || (pc != c.size() && c[pc] < b[pb])) {
      if (a.size() > 0) {
        if (c[pc] == a[a.size() - 1]) {
          pc++;
          mo++;
          continue;
        }
      }
      a.push_back(c[pc++]);
    } else {
      if (a.size() > 0) {
        if (b[pb] == a[a.size() - 1]) {
          pb++;
          mo++;
          continue;
        }
      }
      a.push_back(b[pb++]);
    }
  }
}
vector<int> solve(int x, int y) {
  if (D[y] > D[x]) swap(x, y);
  vector<int> r = c[x];
  gm(r, c[y]);
  for (int i = 17; i >= 0; i--) {
    if (D[x] - D[y] >= (1 << i)) {
      gm(r, BM[x][i]);
      x = T[x][i];
    }
  }
  if (x == y) return r;
  for (int i = 17; i >= 0; i--) {
    if (T[x][i] != T[y][i]) {
      gm(r, BM[x][i]);
      gm(r, BM[y][i]);
      x = T[x][i];
      y = T[y][i];
    }
  }
  gm(r, BM[x][0]);
  return r;
}
int main() {
  int n, m, q;
  scanf("%d%d%d", &n, &m, &q);
  D = vector<int>(n);
  T = vector<vector<int> >(n, vector<int>(18, -1));
  BM = vector<vector<vector<int> > >(n, vector<vector<int> >(18));
  c = vector<vector<int> >(n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--;
    v--;
    cn[u].push_back(v);
    cn[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int t;
    scanf("%d", &t);
    t--;
    if (c[t].size() < 10) c[t].push_back(i);
  }
  dfs(0);
  for (int j = 1; j < 18; j++) {
    for (int i = 0; i < n; i++) {
      int s = T[i][j - 1];
      if (s == -1) continue;
      BM[i][j] = BM[i][j - 1];
      gm(BM[i][j], BM[s][j - 1]);
      T[i][j] = T[s][j - 1];
    }
  }
  for (int i = 0; i < q; i++) {
    int u, v, a;
    scanf("%d%d%d", &u, &v, &a);
    u--;
    v--;
    vector<int> anw = solve(u, v);
    int t = min(int(anw.size()), a);
    printf("%d", t);
    for (int j = 0; j < anw.size(); j++) {
      printf(" %d", anw[j]);
      t--;
      if (t == 0) break;
    }
    printf("\n");
  }
}